Question

In $△$PQR, PQ = 12 cm and PR = 13 cm. $\angle Q={90}^{\circ }$. Find:
tan P - cot R

• A. $0$
• B. $\frac{119}{60}$
• C. $-\frac{119}{60}$
• D. $1$

Answer 1

The correct option is A $0$


Given that, in $△$ PQR, PQ = 12 cm and PR = 13 cm.

Now, from Pythagoras' theorem,
${PQ}^2+{QR}^2={PR}^2$
$\Rightarrow {QR}^2={PR}^2-{PQ}^2$
$\Rightarrow {QR}^2={13}^2-{12}^2$
$\Rightarrow {QR}^2=169-144=25$
$\Rightarrow QR=\sqrt{25}=5\text{cm}$

Now, $tanP=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{QR}{PQ}=\frac{5}{12}$
$cotR=\frac{\text{adjacent side}}{\text{opposite side}}=\frac{QR}{PQ}=\frac{5}{12}$

$\therefore tanP-cotR=\frac{5}{12}-\frac{5}{12}=0$